Partial derivative of integral: Leibniz rule?

The Leibniz rule is as follows:

$$\frac{d}{d\alpha} \int_{a(\alpha)}^{b(\alpha)} f(x, \alpha) dx = \frac{db(\alpha)}{d\alpha} f(b(\alpha), \alpha) - \frac{da(\alpha)}{d\alpha} f(a(\alpha), \alpha) + \int^{b(\alpha)}_{a(\alpha)} \frac{\partial}{\partial\alpha} f(x, \alpha) dx$$

What I would like to know is how to apply the above formula for the case of the partial derivative:

$$\frac{\partial}{\partial\alpha} \int_{a(\alpha)}^{b(\beta)} f(x, \alpha) dx.$$

Solution 1:

You compute a partial derivative with respect to $\alpha$ by holding $\beta$ fixed, and then just differentiating the resulting function of $\alpha$, which is a function of a single variable. And yes, the Leibniz rule tells you how to differentiate this function of $\alpha$.

For a given $\beta$, the derivative of the function \begin{align*} g(\alpha) &= \int_{a(\alpha)}^{b(\beta)} f(x,\alpha) \, dx \end{align*} is \begin{equation} \frac{dg(\alpha)}{d\alpha} = 0 - \frac{da(\alpha)}{d\alpha} f(a(\alpha),\alpha) + \int_{a(\alpha)}^{b(\beta)} \frac{\partial}{\partial \alpha} f(x,\alpha) \, dx. \end{equation}